The maximum velocity of a particle,executing $S$.$H$.$M$. with an amplitude $7 \ mm$ is $4.4 \ ms^{-1}$. The period of oscillation is $\left[\pi=\frac{22}{7}\right]$. (in $s$)

The velocity-time diagram of a harmonic oscillator is shown in the figure. The frequency of oscillation is ..... $Hz$.

The equations for the displacements of two particles in simple harmonic motion are $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$ respectively. The phase difference between the velocities of the two particles at a time $t=0$ is

$A$ particle is performing simple harmonic motion with angular frequency $\omega$ and amplitude $A$. If $a$ is acceleration and $v$ is speed at any instant,then the graph showing the correct variation between $v^2$ and $x^2$ is (where $x$ is displacement):

$A$ particle is performing $S.H.M.$ with a maximum velocity $V$. If the amplitude is doubled and the periodic time is reduced to $\left(\frac{1}{3}\right)^{\text{rd}}$ of its original value,then the new maximum velocity is:

$A$ spring executes $SHM$ with a mass of $10\,kg$ attached to it. The force constant of the spring is $10\,N/m$. If at any instant its velocity is $40\,cm/s$,the displacement will be .... $m$ (where amplitude is $0.5\,m$).